1. Davis-Putnam Methods Computational Logic Lecture 5
Michael Genesereth Autumn 2011

2. Level Saturation

3. Level Saturation Method
function lsm () ;  is a linked list of clauses
{var results  ;
while ({}  )
{results  lss(,results)
  concat(,results)
function lss (, )
{var results  [];
for (var   )
{for (  )
{results  concat(results,resolvents(,))}};
return results}

4. Example for DP
Multiple proofs. Resolutions between parents and children.

5. Davis Putnam Procedure
function dp ()
{for  in vocabulary()
do {var ’{};
for 1 in  for 2 in 
such that   1   2
do {var ’ 1 {}  2  {};
if not tautology(’) then ’’{’}};
  { |    or   }  ’};
return {if {}   then unsatisfiable else satisfiable}}
function tautology()
{  and   }

6. Example Without DP {p, q, r} {q, r} {p} {p, q, r} {q, r} {p}
{p, q, r} {q, r} {q}
{p, q, r} {q, r} {q}
{p, q, r} {p, r} {r}
{p, q, r} {p, r} {r}
{p, q, r} {p, r}
{p, q, r} {p, r} {}
{p, q}
{p, q} Cost =
{p, q} 378 resolutions
{p, q}

7. Example With DP {p, q, r} {q, r} {p, q, r} {q, r} {p, q, r} {q, r}
{p, q, r} {r}
{p, q, r} {r}
{p, q, r} {}
Cost = = 21 resolutions

8. Motivation for DPLL
DP can cause a quadratic expansion every time it is applied. This can easily exhaust space on large problems.
DPLL attacks this problem by sequentially solving smaller problems.
Basic idea: Choose a literal. Assume true, simplify clause set, and try to show satisfiable. Repeat for the negation of the literal. Good because we do not cross multiply the clause set.

9. Davis Putnam Logemann Loveland
function dpll ()
{var ;
if  = {} then return yes;
if {}  then return no;
  choose vocabulary());
if dpll(simplify(, )) return yes
else return dpll(simplify(,))}
function simplify (, )
{var ’;
for   
do {if  then skip
else if negation()
then ’ ’  {  {negation()}}
else ’ ’  {}}}

10. Simplification Example
Clauses:
{p,q}
{p,r}
{r,s}
Literal: p
Simplification:
{r}

11. Comparisons Problem Tautology DP DPLL Prime 30.00 0.00 0.00
Prime16 > 1 hour * 9.15
Prime17 > 1 hour * Mkadder32 >> 1 hour
Mkadder42 >> 1 hour
Mkadder52 >> 1 hour
Mkadder53 >> 1 hour
Mkadder63 >> 1 hour *
Mkadder73 >> 1 hour *

12. Coin Logic Syntax: The logical constants: quarters, nickels, dimes
Negation: coin upside down
Disjunction: stack of coins
Conjunction: set of stacks
Almost exactly clausal form.
Propositional Resolution:
Add two stacks,
deleting at most one pair of complementary coins.

13. Example Quarter means it is Monday. Nickel means Mary loves Pat.
Dime means Mary loves Quincy.
[Nickel,Dime]: If Mary loves Pat, Mary loves Quincy.
[Quarter,Nickel,Dime]: Monday Mary loves P or Q.
[Nickel, Dime]:Mary loves only one of the two.
[Quarter,Dime]: If Monday, Mary loves Quincy.
[Quarter,-Nickel]: If Monday, Mary does not love Pat.